Problem: The equation of hyperbola $H$ is $\dfrac {(y+7)^{2}}{49}-\dfrac {(x-6)^{2}}{9} = 1$. What are the asymptotes?
Answer: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+7)^{2}}{49} = 1 + \dfrac {(x-6)^{2}}{9}$ Multiply both sides of the equation by $49$ $(y+7)^{2} = { 49 + \dfrac{ (x-6)^{2} \cdot 49 }{9}}$ Take the square root of both sides. $\sqrt{(y+7)^{2}} = \pm \sqrt { 49 + \dfrac{ (x-6)^{2} \cdot 49 }{9}}$ $ y + 7 = \pm \sqrt { 49 + \dfrac{ (x-6)^{2} \cdot 49 }{9}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 7 \approx \pm \sqrt {\dfrac{ (x-6)^{2} \cdot 49 }{9}}$ $y + 7 \approx \pm \left(\dfrac{7 \cdot (x - 6)}{3}\right)$ Subtract $7$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{7}{3}(x - 6) -7$